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This question already has an answer here:

Let $H\triangleleft G$, and H is a subgroup of G such that $[G:H] = k$, where $[G:H]$ states the number of left cosets of $H$ in $G$.

1) How may I show that for all $a\in G, a^k\in H$?

2) If we do NOT assume the normality of $H$, can I have any counterexample showing that 1) is not true?

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marked as duplicate by Watson, Dietrich Burde, Community Nov 3 '16 at 16:52

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  • $\begingroup$ Oh, I meant "of the index $[G:H]$". $\endgroup$ – omosiroiP. Nov 3 '16 at 16:18
  • $\begingroup$ One typically says "is a subgroup of index $k$", without the "the". $\endgroup$ – Daniel Fischer Nov 3 '16 at 16:25
  • $\begingroup$ @DietrichBurde The last correction may help you clarify...? $\endgroup$ – omosiroiP. Nov 3 '16 at 16:25
  • $\begingroup$ Oh, I got it....@DanielFischer $\endgroup$ – omosiroiP. Nov 3 '16 at 16:25
  • $\begingroup$ Concerning question 2), look at the smallest group having a non-normal subgroup. Look at a non-normal subgroup of that. $\endgroup$ – Daniel Fischer Nov 3 '16 at 16:25
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I assume, $[G:H]=k$, then the order of $G/H$ is $k$ and $\bar a^k=1$ where $\bar a$ is the class of $a$ in $G/H$ and $a^k\in H$, since its class in $G/H$ is $1$.

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